By A. T. Berztiss and Werner Rheinboldt (Auth.)

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We shall consider two cases. (a) x = y. Then (ii) reduces to {{x}}. Since by hypothesis, and must also be a singleton. This means that {a ft}must equal {a} giving a = ft. Hence (i) reduces to {{a}} and, since (i) and (ii) are equal by hypothesis, {{a}} = {{*}}, giving a = x. It follows that all a ft,x, and y are equal, in particular a = x andft= j . (b) x ^ y. Then (ii) contains exactly one singleton and exactly one pair. Since ft> = (x, y} (i) must contain one singleton and one pair.

1 (a) In which instances are the sets A and B equal? (i) A = {1, 2, 3}, B = {i 11 is an integer}. (ii) A = {1, 2, 3, 4, 5}, B = {1, 10, 11, 100, 101}. (iii) A = {a,P, y} B={a,b,c}. (iv) A = {a}, B = {x I x is the first letter of the Greek alphabet}. (v) A = {a, b, ab, ba b, a} B = {a, b, ab}. 3 by induction. (Consider a statement P(ri) about some nonnegative integer n. The principle of induction states: If P(k) is true and P(m) implies P(m + 1), then P(n) is true for all n such that k ^ n ^ m + 1.

3 The system <{0, 1}, ©, * , ' , 0, 1> is a Boolean algebra. Proof. The given system is precisely the Boolean algebra ©, *, ', 0, 1> of Section le, with set B having just two elements, the distinct elements 0 and 1. To prove that the system is a Boolean algebra it suffices to show that {0, 1} is closed under ©, *, and '. 12 (all of Section le) we have the following definitions of operations ©, *, ' for members of {0, 1}: 0 O ©© 1I O © O = O, 0*0 = 0 * 1 == = O' = 1 © I<3 = 1 * 0 = 1, 0 = 0, 0, 1 0 1 = 1; 1 * 1 = 1; R = O.

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